Question

Butane gas is used as a lighter fuel for a common lighter orbutane torch.What volume (in L) will a 32 g sample of butane gas,C4H10(9), occupy at a temperature of 45.0 °C and apressure of 728 mm Hg?

Answer 1

1) List known values

Mass: 32g (C4H10)

Temperature: 45.0ºC

Pressure: 728 mmHg

Ideal constant: 0.082057 L⋅atm⋅K−1⋅mol−1

List unknown values

Volume:

2) Convert grams of C4H10 into moles of C4H10

The molar mass ops C4H10 is 58.1222 g/mol

$\text{molesofC}_4H_{10}=32gofC_4H_{10}\cdot\frac{1molofC_4H_{10}}{58.1222gC_4H_{10}}=0.5506molofC_4H_{10}$

3) Convert the temperature from ºC to K

K = 273.15 K + 45.0ºC = 318.15 K

4) Convert pressure from mmHg to atm

$\text{atm}=728\operatorname{mm}Hg\cdot\frac{1\text{ atm}}{760\text{ mmHg}}=0.9579\text{atm}$

5) Set the equation and solve for V

$\text{PV}=\text{nRT}$$\frac{PV}{P}=\frac{nRT}{P}$$V=\frac{nRT}{P}$

6) Plug in known quantities

n: 0.5506 mol

R: 0.082057 L⋅atm⋅K−1⋅mol−1

T: 318.15 K

P: 0.9579 atm

$V=\frac{(0.5506\text{mol)}(0.082057L\cdot\text{atm}\cdot K^{-1}mol^{-1})(318.15\text{ K)}}{0.9579atm}=\text{ 15.00 L}$

Butane gas will occupy a volume of 15.00 L in the conditions described.

7) Find the volume occupy by butane using R: 8.314 L⋅kPa⋅K−1⋅mol−1

Convert pressure from mmHg to kPa

$\text{kPa}=728\operatorname{mm}Hg\cdot\frac{101325Pa}{760\operatorname{mm}Hg}\cdot\frac{1\text{kPa}}{1000Pa}=97.0587\text{kPa}$

Plug in values

n: 0.5506 mol

R: 8.314 L⋅kPa⋅K−1⋅mol−1

T: 318.15 K

P: 97.0587kPa

$V=\frac{(0.5506mol)(8.314L\cdot kPa\cdot K^{-1}mol^{-1})(318.15\text{ K)}}{97.0587\text{ kPa}}=\text{ 15.01 L}$